STATISTICS 402B
Spring 2016
Homework Set#8 Solution
1. .
(a) Aliasing
A =
B =
C =
D =
E =
Pattern
BCDE
ACDE
ABDE
ABCE
ABCD
AB
AC
AD
AE
BC
BD
BE
CD
CE
DE
=
=
=
=
=
=
=
=
=
=
CDE
BDE
BCE
BCD
ADE
ACE
ACD
ABE
ABD
ABC
(b) Data Matrix
A
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
B
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
C
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
D
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
E
1
-1
-1
1
-1
1
1
-1
-1
1
1
-1
1
-1
-1
1
(c) .
Term
Effect Estimate
A
-0.25
B
7.25
C
5.85
D
1.45
E
-0.95
A*B
0.95
A*C
-1.05
A*D
-0.05
A*E
-0.55
B*C
2.75
B*D
-0.15
B*E
-0.35
C*D
0.15
C*E
-0.35
D*E
-0.45
1
Y
14.8
14.5
18.1
19.4
18.4
15.7
27.3
28.2
16
15.1
18.9
22
19.8
18.9
29.9
27.4
(d) .
(e) The important effects are main effects B, C and D and the interactions AB and BC. We may omit
the AB interaction on account of the fact that A is a negligible effect (implying weak heredity. Effects
that need interpretation are the main effect D by itself and the interaction BC.
The main effect D is to increase the mean of Y by 1.45 units. Compared with e range of values for Y
(14.5 to 29.9) this can be considered insignificant.
The effect of C is larger at the higher level of B and is to increase the mean of Y by 10 units, and can
be considered a signifcant increase.
2
(f) Resolution is V (the length of the word in the genenerator is 5). It tells you that in this design the
main effects are NOT aliased with either 2-fi or 3-fi and that the 2-fi are aliased with 3-fi’s or higher
order interactions only. Thus if the 3-fi and higher order are assumed to be negligible we may estimate
the main effects and 2-fi’s by fitting a 2nd degree factorial model.
2. . Defining relation: I=ABCE=BCDF=ADEF The alias structure is:
AB = CE
AC = BE
AD = EF AE = BC = DF
AF = DE
BD = CF
BF = CD
run
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Basic Design
A B C D
+ - + + + - + + - + - + + + + + - +
+ - +
- + - +
+ + - +
- + +
+ - + +
- + + +
+ + + +
E=ABC
+
+
+
+
+
+
+
+
F=BCD
+
+
+
+
+
+
+
+
(1)
ae
bef
abf
cef
acf
bc
abce
df
adef
bde
abd
cde
acd
bcdf
abcdef
3. .
(a) I=ABD=ACE=BCF=BCD=ACD=ABEF=DEF
(b) Derive the
A
(1) a
+
b
ab +
c
ac +
bc
abc +
treatment combinations to be used in this experiment:
B C D=AB E=AC F=BC
+
+
+
def
+
af
+ +
be
+ +
abd
- +
+
cd
- +
+
ace
+ +
+
bcf
+ +
+
+
+
abcdef
(c) Because 6 factors are investigated in 8 runs, k 6= N − 1. Not saturated.
(d) Resolution is III
(e) Give the alias pattern of the main effects and 2-factor interactions:
A+BD+CE
B+AD+CF
C+AE+BF
D+AB+EF
E+AC+DF
F+BC+DE
3
(f) Another 1/8th
A B C
+ + +
- + +
+ - +
- +
+ + - + + -
fraction
D E
+ +
+ - +
- +
+ + +
-
was added. Show the treatment combinations used in this design:
F Runs
abc
- bcde
+ acdf
+
cef
+ abef
+ bdf
ade
(1)
(g) I=-ABD=-ACE=-BCF
A-BD-CE
B-AD-CF
C-AE-BF
D-AB-EF
E-AC-DF
F-BC-DE
(h) Give the alias structure of the combined fraction (again assuming 3-factor and higher order interactions
are negligible)
A BD+CE
B AD+CF
C AE+BF
D AB+EF
E AC+DF
F BC+DE
(i) IV
4